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Any matrix $A$ can be presented as a sum of its symmetrical and skew-symmetrical part:

$A=sym(A)+skew(A)$.

Decomposition can go further and we can present symmetrical part as a sum of some diagonal matrix and hollow (zeros on diagonal) symmetrical matrix. Now we have (used below notation is only for the needs of this question)

$A= diag(A)+hols(A)+ skew(A)$.

Let's assume the dimension of $A$ is $3$.

In this case we have the sum of $3$ components where every component has $3$ DOF (decoded in three 3D vectors). We can also present these components in the form

$A= k_d{diag_n}(A )+k_h{hols_n}(A )+ k_s{skew_n}(A )$,

where coefficients

$k_d, k_h, k_s$

are calculated so to assure that vectors which represent DOF of components are unit vectors, index $n$ denotes here this kind of "normalization" (so in these unit vectors we have decoded 2 degrees of freedom and additional DOF is in the appropriate scaling coefficient $k_{\{d,h,s)\}}$)

Example of such decomposition:

$\begin{bmatrix} 1 & 3 & 5 \\ 1 & 4 & 4 \\ 1 & 8 & 8 \end{bmatrix}$ = 9$\begin{bmatrix} \dfrac{1}{9} & 0 & 0 \\ 0 & \dfrac{4}{9} & 0 \\ 0 &0 & \dfrac{8}{9} \end{bmatrix}$+ 7$\begin{bmatrix} 0 & \dfrac{2}{7} & \dfrac{3}{7} \\ \dfrac{2}{7} & 0 & \dfrac{6}{7} \\ \dfrac{3}{7} & \dfrac{6}{7} & 0 \end{bmatrix}$+ 3$\begin{bmatrix} 0 & \dfrac{1}{3} & \dfrac{2}{3} \\ -\dfrac{1}{3} & 0 & -\dfrac{2}{3} \\ -\dfrac{2}{3} & \dfrac{2}{3} & 0 \end{bmatrix}$

Two of components have distinct geometric interpretation:
$diag(A)$ is simply a scaling matrix and $skew(A)$ is a scaled composition of a projection and rotation by $\pi/2$ (See question*).
However the geometric interpretation for the $hols_n(A)$ is unknown to me..
In the general case matrix $hols(A)$ has a form

$\begin{bmatrix} 0 & a & b \\ a & 0 & c \\ b & c & 0 \end{bmatrix}$

Some properties of this matrix maybe provide some information about its nature:

  • it has real eigenvalues (as a special case of symmetric matrix) so rotation probably can't be engaged for this interpretation (unless it is by $\pi$)
  • determinant of this matrix $det(hols(A))= 2abc$ so its rank $=3$ if $ a,b,c \neq 0$. If one of $a,b,c=$ $0$ its rank decreases
    (unlike $skew_n(A)$ where zeroing a single entry (and its skew-symmetrical one) doesn't change geometric interpretation - only axis of rotation which determines at the same time the direction of projection).

  • few others are in Wikipedia.

So my question is:

  • Could hollow symmetrical matrix $hols_n(A)$ be decomposed in such a way that geometric interpretation of $hols_n(A)$ would be explicit ?

If geometric interpretation of the hollow matrix is too difficult one can propose other method for decomposition of symmetrical part $sym(A)$ only on condition that both parts should have 3 DOF and each part should have explicit geometric interpretation..

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  • $\begingroup$ The diagonal-hollow decomposition is not basis independent. When you extract the hollow part, you have chosen a coordinate frame. If the hollow part has any geometric meaning, its significance may be limited to this particular coordinate frame only. $\endgroup$ – user1551 Jul 14 '16 at 11:22
  • $\begingroup$ @user1551 hmm, what about a standard frame? $\endgroup$ – Widawensen Jul 14 '16 at 11:25
  • $\begingroup$ @usser1 eigenvalues for this part would be the same after the change of frame? If so properties of this matrix would be somehow preserved.. $\endgroup$ – Widawensen Jul 14 '16 at 11:42
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    $\begingroup$ The obvious property is that each basis vector is mapped to a vector orthogonal to it. $\endgroup$ – celtschk Jul 14 '16 at 15:19
  • $\begingroup$ Ok, could we extend this property somehow for transformations, for example, of cubes with unit sides? $\endgroup$ – Widawensen Jul 14 '16 at 15:22
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Just as a note aside, if instead of making it hollow, you fill the diagonal with $f_{1,\,2,\,3} (a,b,c)$ then you can get some significant factorization. For example: $$ \left( {\begin{array}{*{20}c} 1 & a & b \\ a & {1 + a^2 } & c \\ b & c & {1 + b^2 + (c - ab)^2 } \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} 1 & 0 & 0 \\ a & 1 & 0 \\ b & {c - ab} & 1 \\ \end{array} } \right)\left( {\begin{array}{*{20}c} 1 & a & b \\ 0 & 1 & {c - ab} \\ 0 & 0 & 1 \\ \end{array} } \right) $$

Now, doing that, the 3 DOF are preserved ; but, taking them out of the diagonal matrix, the DOF there get increased.
I presume it is related to @user1551's note, but there is much to dig about..

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  • $\begingroup$ Interesting symmetry in your decomposition, you are right, maybe others forms of matrices for decomposition of symmetrical part are more interesting (on the condition that we have 3 DOF preserved). Complementing diagonal form always is easy for interpretation.. just particular kind of scaling ... how would you be interpreting geometrically these triangular matrices above ? $\endgroup$ – Widawensen Jul 14 '16 at 15:47
  • $\begingroup$ @Widawensen, that was just the first "interesting object" coming out from a very first digging ! then, by e.g. putting $c=a(b+1)$ it simplifies further, and acquire a more clear meaning, but .. ODF lower to 2. So a very interesting subject, with many facets to explore: thanks for proposing. We'll keep in touch for the outcomings. $\endgroup$ – G Cab Jul 14 '16 at 16:08
  • $\begingroup$ Can we regard these 3 matrices in decomposition as some kind of "basis" for 3D matrix ? Family of 3D matrices can be composed from them just as a linear combination.. $\endgroup$ – Widawensen Jul 14 '16 at 16:24
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What you are proposing is and additive transformation of coordinates.

$$ \mathbf{A} = diag\left( \mathbf{A} \right) + hollow\left( \mathbf{A} \right) $$ corresponds to that
the new vector $\mathbf{x}' = \left( {x'_1 ,x'_2 ,x'_3 } \right)$ is built first by $\left( {x'_1 ,0 ,0 } \right)$ (which is parallel to $\mathbf{x}$ ) plus $ \left( {0 ,x'_2 ,x'_3 } \right)$ (which is normal to $\mathbf{x}$ and to the previous step). Same for $\mathbf{y}'$ and $\mathbf{z}'$.
So a general Hollow matrix brings each base vector onto the plane normal to it. That is a rotation of $\pi/2$ around an axis normal to $\mathbf{x}$ ($1$ DOF) and a stretch ($1$ DOF): tot. $6$ DOF.
Symmetric and Skew-symmetric components do the same projections of each axis onto a plane normal to it, but the DOFs split into $3$ per each component, due to interrelation among rotation axes and stretch factors.
Of course the interrelation is different between the two components, but in this interpretation I cannot see a clearcut among the two.

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  • $\begingroup$ However I would not be touching the skew-symmetric part because it has good geometrical interpretation. In the light of this question it would be interesting to think over what means for 3D matrix to have geometric interpretation - some matrices evidently have them easier to find, others are hard for finding. Maybe it is really some kind of condition which gives such possibility. Whatever it is, hollow symmetric has 3DOF and it is one of the simplest forms of this kind. $\endgroup$ – Widawensen Jul 15 '16 at 5:22

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